What is the ratio of kinetic energy of 7 g o | vedclass

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(B) The average kinetic energy of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Since the temperature $T$ is constant, the kinetic ene

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$2 : 1$

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(B) The average kinetic energy of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Since the temperature $T$ is constant, the kinetic energy is directly proportional to the number of moles $n$: $KE \propto n$. Calculate the moles of $N_2$: $n_{N_2} = \frac{7 \ g}{28 \ g/mol} = 0.25 \ mol$. Calculate the moles of $O_2$: $n_{O_2} = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$. The ratio of kinetic energies is: $\frac{(KE)_{N_2}}{(KE)_{O_2}} = \frac{n_{N_2}}{n_{O_2}} = \frac{0.25}{0.125} = \frac{2}{1}$. Therefore, the ratio is $2 : 1$.

What is the ratio of kinetic energy of $7 \ g$ of nitrogen and $4 \ g$ of oxygen at $T \ K$?

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